Given the definition of weighted matrix max-norm as
\begin{gather*} \left \| A \right \| _\infty ^w = \max \limits _{x \ne 0} \frac{\left \| A x\right \| _\infty ^w}{\left \| x\right \| _\infty ^w} \tag{1} \label{eq:special} \end{gather*}
and the definition of weighted vector max-norm as
\begin{gather*}
\left \| x\right \| _\infty ^w = \max \limits _{i} |\frac{x_i}{w_i}|, w_i>0. \tag{2} \label{eq:special1}
\end{gather*}
I'm trying to prove there's an equivalent representation
\begin{gather*}\left \| A \right \| _\infty ^w = \left \| |A|w \right \| _\infty ^w \tag{3} \label{eq:special2} \end{gather*}
but with no luck. I managed to prove $\left \| A \right \| _\infty := \max \limits _{x \ne 0} \frac{\left \| A x\right \| _\infty }{\left \| x\right \| _\infty } = \max \limits _i \sum \limits _j |a_{ij}|$, which seems related but didn't really help. Can you show me how to work it out? Thanks in advance.
Let $W$ denote the matrix $\operatorname{diag}(w)$. Note that $\|x\|_\infty^w = \|W^{-1}x\|$. It follows that $$ \begin{align} \|A\|_\infty^w &= \max_{x \neq 0}\frac{\|Ax\|_\infty^w}{\|x\|_\infty^w} = \max_{x \neq 0}\frac{\|W^{-1}Ax\|_\infty}{\|W^{-1}x\|_\infty} = \max_{y \neq 0} \frac{\|W^{-1}A(Wy)\|_\infty}{\|W^{-1}(Wy)\|_\infty} \\ & = \max_{y \neq 0} \frac{\|(W^{-1}AW)y\|_\infty}{\|y\|_\infty} = \|W^{-1}AW\|_\infty. \end{align} $$ Combining this with your characterization of $\|A\|_\infty$, we have $$ \|A\|_\infty^w = \max_i \sum_j \frac{w_j}{w_i}|a_{ij}| = \max_i \frac 1{w_i} (e_i^T|A|)w = \big\||A|w \big\|_\infty^w. $$ In the above, $e_i$ denotes the $i$th standard basis vector )i.e. the $i$th column of the identity matrix) and $|A|$ denotes the matrix whose entries are $|a_{ij}|$.